Question 30
(a)
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(b)
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(c)
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BC = 8 sin 18° = 2.5 ms-1
BD = 5.0 ms-1 in direction W18° S
Question 31
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(a) FH = 100 sin 60° = 87 N
(b) FV = 100 cos 60° = 50N
Question 32
vPG = 250/0.5 = 500 km-1 E
vAG = 50 kmh-1 N (not S); hence vGA = 50 kmh-1 S
vPA = vPG + vGA = 500 kmh-1 E + 50 kmh-1 S
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vPA = 502.5 kmh-1 ; angle q = tan-1 50/500 = 5.7° ; hence direction is E 5.7° S
Question 33
dP = 4 dQ and rP = 4 rQ
(a) AP = 4p rP2
AQ = 4p (rP/4)2 = 4p rP2 /16 = AQ/16
Hence AP = AQ/16
(b) VP = 4/3 p rP3
VQ = 4/3 p rQ (1/4)3
Hence VP = VQ/64
(c) mP = r VP
mQ = r VQ
mP = mQ/64
(d) r P = r Q
Question 34
(a) V1 = 35 V2
4/3 p r13 = 35 x 4/3 p r22
(b)
Square both sides of the equation to give: Z2 = R2 + 4p 2L2f2
Rearrange the equation to: Z2 = 4p 2L2f2 + R2
This is now in the form y = mx + c
where the slope (m) = 4p 2L2 and the intercept on the y-axis (c) = R2.
Plot the graph:![]()
Slope = (44100 - 10000)/90000 = 0.379
4p 2L2 = 0.379
L = 0.098 metres
Intercept = 10000 = R2
Therefore R = 100 ohm
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